利用ES6中的Array.find/ Array.findIndex来判断数组中已存在某个对象

【利用ES6中的Array.find/ Array.findIndex来判断数组中已存在某个对象】的更多相关文章

利用ES6中的Array.find/ Array.findIndex来判断数组中已存在某个对象

前端开发过程中,我们会经常遇到这样的情景:比如选中某个指标obj,将其加入到数组checkedArr中({id: 1234, name: 'zzz', ...}),但是在将其选中之前要校验该指标是否已经被选择. 以前的思路是:循环数组checkedArr,如果checkedArr[i].id===obj.id,则说明该指标已经在数组中了. 在ES6中,数组得到扩展,新增了find和findIndex两个方法,可以用到这个情境中: find() 方法返回数组中满足提供的测试函数的第一个元素的值.…

[LeetCode] Search in Rotated Sorted Array II 在旋转有序数组中搜索之二

Follow up for "Search in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the array. 这道是之前那道Search in Rotated Sorted Array 在旋转有序数组…

【LeetCode】Find Minimum in Rotated Sorted Array 找到旋转后有序数组中的最小值

本文为大便一箩筐的原创内容,转载请注明出处,谢谢:http://www.cnblogs.com/dbylk/p/4032570.html 原题: Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). Find the minimum element. You may assume no duplica…

[LeetCode] 81. Search in Rotated Sorted Array II 在旋转有序数组中搜索之二

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]). You are given a target value to search. If found in the array return true, otherwise return false. Ex…

LeetCode 80 Remove Duplicates from Sorted Array II（移除数组中出现两次以上的元素）

题目链接:https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/#/description 给定一个已经排好序的数组,数组中元素存在重复,如果允许一个元素最多可出现两次,求出剔除重复元素(出现次数是两次以上的)后的数组的长度. For example, Given sorted array nums = [1,1,1,2,2,3], Your function should return length = 5,…

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Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memo…